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Consider the plane (P,L) where P = {1, 2, 3, a, b, c} L = {{1, 2, 3} , {a, b, c}} {{i, j} i

Consider the plane (P,L)

where P = {1, 2, 3, a, b, c}

L = {{1, 2, 3} , {a, b, c}} {{i, j} i = 1, 2, 3, j = a, b, c}

(assuming that a, b, c, 1, 2, 3 are pairwise different). In order to define a plane with segments (P,L, ) having this as underlying plane, we only need to specify total orders {1,2,3} and {a,b,c} because the other elements of L have only 2 elements and therefore a unique pair of mutually reverse total orders.

Is there a order function so that (P,L, ) is a plane with segments?

Hint: we may assume that 2 is between 1, 3 and b is between a, c (why?). Look at the half planes of the line (2b). Show that 1 and 3 must be in the same half plane, and show that they must be in different half planes. A contradiction.

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